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3 years ago

Select all of the objects Mae will need to use in her experiment.

Physics

2 answers:

balandron [24]3 years ago

5 0

Answer: magnet,light bulb and wire coil

Explanation:

Bond [772]3 years ago

4 0

She’ll need all objects to make an electrical current inside a magnetic field

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What does the solar system consist of?

scZoUnD [109]

Planets, Sun, stars, plasma, and black matter

8 0

4 years ago

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What are the horizontal and vertical velocities of a stunt bike that leaves a ramp at 100 km/hr and at an angle of 35 degrees?

Alex787 [66]

Horizontal velocity: 81.9 km/h

Vertical velocity: 57.4 km/h

Explanation:

We can solve this problem by resolving the velocity vector into its component along the horizontal and vertical direction.

The horizontal velocity of the stunt bike is given by:

$v_x = v cos \theta$

where

v = 100 km/h is the magnitude of the velocity

$\theta=35^{\circ}$ is the angle of projection

Substituting, we find

$v_x = (100)(cos 35^{\circ})=81.9 km/h$

The vertical velocity instead is given by

$v_y = v sin \theta$

where

$v=100 km/h$

$\theta=35^{\circ}$

Substituting,

$v_y = (100)(sin 35^{\circ})=57.4 km/h$

Learn more about vector components:

brainly.com/question/2678571

#LearnwithBrainly

4 0

3 years ago

Two carts, one twice as heavy as the other, are at rest on a horizontal track. A person pushes each cart for 8 s. Ignoring frict

GalinKa [24]

Answer:

The correct option is: B that is 1/2 K

Explanation:

Given:

Two carts of different masses, same force were applied for same duration of time.

Mass of the lighter cart = $m$

Mass of the heavier cart = $2m$

We have to find the relationship between their kinetic energy:

Let the KE of cart having mass m be "K".

and KE of cart having mass m be "K1".

As it is given regarding Force and time so we have to bring in picture the concept of momentum Δp and find a relation with KE.

Numerical analysis.

⇒ $KE = \frac{mv^2}{2}$

⇒ $KE = \frac{mv^2}{2}\times \frac{m}{m}$

⇒ $KE = \frac{m^2v^2}{2}\times \frac{1}{m}$

⇒ $KE = \frac{(mv)^2}{2}\times \frac{1}{m}$

⇒ $KE = \frac{(\triangle p)^2}{2}\times \frac{1}{m}$

⇒ $KE = \frac{(\triangle p)^2}{2m}=\frac{(F\times t)^2}{2m}$

Now,

Kinetic energies and their ratios in terms of momentum or impulse.

KE (K) of mass m.

⇒ $K=\frac{(F\times t)^2}{2m}$ ...equation (i)

KE (K1) of mass 2m.

⇒ $K_1=\frac{(F\times t)^2}{2\times 2m}$

⇒ $K_1=\frac{(F\times t)^2}{4m}$ ...equation (ii)

Lets divide K1 and K to find the relationship between the two carts's KE.

⇒ $\frac{K_1}{K} =\frac{(F\times t)^2}{4m} \times \frac{2m}{(F\times t)^2}$

⇒ $\frac{K_1}{K} =\frac{2m}{4m}$

⇒ $\frac{K_1}{K} =\frac{2}{4}$

⇒ $\frac{K_1}{K} =\frac{1}{2}$

⇒ $K_1=\frac{K}{2}$

⇒ $K_1=\frac{1}{2}K$

The kinetic energy of the heavy cart after the push compared to the kinetic energy of the light cart is 1/2 K.

7 0

3 years ago

Which form of energy body will possess when it is kept on the top of Hill

Oduvanchick [21]

Potential energy would be the answer

3 0

3 years ago

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Consider the following distinct forces:______________________________.

Kamila [148]

Answer:

1 and 2

Explanation:

It is given that, force exerted by air is negligible in any way.

Also, it is given that channel is in the shape of a segment of a circle with its center at O.

When it is within the frictionless channel at position 'Q',

A gravity exerts force on the ball in the downward direction. On the other hand, the channel pointing from Q to O also exerts a force on the ball.

However, there is no any force in the direction of motion. On the other hand, he channel pointing from O to Q does not exert a force on the ball.

5 0

4 years ago